Proof of Inequality using AM-GM I just started doing AM-GM inequalities for the first time about two hours ago. In those two hours, I have completed exactly two problems. I am stuck on this third one! Here is the problem:
If $a, b, c \gt 0$ prove that $$ a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a.$$
I am going crazy over this! A hint or proof would be much appreciated. Also any general advice for proving AM-GM inequalities would bring me happiness to my heart. Thank you!
 A: Using rearrangement inequality with $a^2,b^2,c^2$ and $a,b,c$ we get
$$aa^2+bb^2+cc^2\ge a^2b +b^2c+c^2a$$
A: You already have a great answer from @Adriano.  To use AM-GM here, in general the idea would be to observe exponents on both sides and try finding a convex combination of $(3, 0, 0), (0, 3, 0)$ and $(0, 0, 3)$ which gives you a term like $(2, 1, 0)$. 
Muirhead's inequality - if you're familiar with it - assures us this will work as $[3, 0, 0] \succ [2, 1, 0]$
So you may consider the following generic equation with non-negative $\alpha+\beta+\gamma=1$:
$$\alpha (3, 0, 0)+\beta (0, 3, 0) +\gamma(0, 0, 3)= (2, 1, 0)$$
Obviously $\alpha = \frac23, \beta = \frac13, \gamma=0$ comes to mind.  Thus the basic inequality to use would be the AM-GM:
$$\tfrac23a^3 + \tfrac13b^3 \ge a^2b$$
Summing three similar inequalities get you the result. 
A: $$\color{blue}{a^2(a-b)+b^2(b-c)+c^2(c-a)=(a^2-c^2)(a-b)+(b^2-c^2)(b-c)}$$
$$\color{red}{a\geq b \geq c \implies (a^2-c^2)(a-b)+(b^2-c^2)(b-c) \geq0}$$ 
A: One trick that shows up a lot with AM-GM inequalities is to apply AM-GM several times, then combine them all together somehow. Let's try that. AM-GM tells us that:
$$
x + y + z \geq 3 \sqrt[3]{xyz} \tag{$\star$}
$$
Now for what values of $x,y,z$ would the LHS and RHS of $(\star)$ match up with some of the terms in the LHS and RHS of the desired inequality? In particular, suppose that we want the cube root to be $a^2b$. Then $xyz = a^6b^3$, which suggests that we take $x = y = a^3$ and $z = b^3$. Repeating this, we obtain:
\begin{align*}
a^3 + a^3 + b^3 &\geq 3 \sqrt[3]{a^3a^3b^3} = 3a^2b \\
b^3 + b^3 + c^3 &\geq 3 \sqrt[3]{b^3b^3c^3} = 3b^2c \\
c^3 + c^3 + a^3 &\geq 3 \sqrt[3]{c^3c^3a^3} = 3c^2a \\
\end{align*}
Adding up the above three inequalities and dividing through by $3$ yields the desired inequality. $~~\blacksquare$
